Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 5 / 1

Use the bilinear pairing $T_g(f)={\int }_{{\mathbb{R}}^n}fg$, so the map $g\mapsto T_g$ is complex linear. With the sesquilinear convention $\int f\overline{g}$ the representing map would instead be conjugate linear. For conjugate exponents $p,q$, Hölder's inequality gives $|T_g(f)|\le \Vert g{\Vert }_q\Vert f{\Vert }_p$. If $g\ne 0$, take $v=|g{|}^{q-2}\overline{g}$, defined as zero where $g=0$. ThenConsequently $\Vert T_g\Vert =\Vert g{\Vert }_q$. This also proves injectivity of the representation.

For surjectivity, let $\ell$ be a nonzero element of the continuous dual space of $L^p$ and choose $f_0$ with $\ell (f_0)=1$. Its kernel $K$ is a closed linear subspace. Use the allowed best-approximation fact to choose $h\in K$ and put $z=f_0-h$. Thus $\ell (z)=1$ andFor each $f$, the vector $f-\ell (f)z$ belongs to $K$. Therefore $\Phi (f)=\ell (f)\Vert z{\Vert }_p^p$, givingMembership follows from $(p-1)q=p$; the zero functional is represented by zero. This proves the isometric isomorphism $(L^p({\mathbb{R}}^n){)}^{\ast }\cong L^q({\mathbb{R}}^n)$ without using a representation theorem as a substitute for the proof. It is the closed-hyperplane proof of Lp duality.

The relevant Banach-Alaoglu theorem says that for $1<p<\infty$ the closed bounded balls of $L^p({\mathbb{R}}^n)$ are compact in the weak topology. In particular, every bounded sequence has a weakly convergent subsequence whose limit has norm no larger than the common bound. Applying the preceding duality of Lp spaces also identifies this as weak-star compactness of $L^p=(L^q{)}^{\ast }$; its weak and weak-star test spaces are both $L^q$.

Here is a direct proof of both compactness and the subsequence assertion. Let $\Vert f_j{\Vert }_p\le M$ and choose a countable dense set $\{g_m\}$ in the separable Lp space $L^q$. Each scalar sequence $\int f_j{g}_m$ is bounded. Successive subsequence selection followed by the diagonal subsequence argument gives $f_{j_k}$ for which these pairings converge for every $m$. For arbitrary $g\in L^q$,First approximate $g$ by $g_m$ and then take $k,l$ large. This proves convergence of every pairing, and the limit $\Lambda (g)={\lim }_k\int f_{j_k}g$ is linear with $|\Lambda (g)|\le M\Vert g{\Vert }_q$. Apply the already proved duality of Lp spaces with $q$ in place of $p$: $\Lambda (g)=\int fg$ for an $f\in L^p$ with $\Vert f{\Vert }_p\le M$. ThusTo obtain compactness rather than only a sequential statement, on the radius-$M$ ball defineDensity and Hölder's inequality show that this is a metric space distance and that its topology is exactly the weak topology on this ball: every remaining test pairing is uniformly approximated there by a dense-set pairing. The subsequence argument proves sequential compactness for this metric. A sequentially compact metric space is compact: otherwise either an infinite separated sequence contradicts total boundedness or a Cauchy sequence without a limit contradicts sequential compactness; completeness and total boundedness give compactness by successive finite coverings. This completes the proof. The same argument is the Sequential Banach-Alaoglu theorem for a separable predual.

The assertion that fails is weak compactness of the $L^1$ unit ball; the abstract Banach-Alaoglu theorem for dual balls never fails. Consider the concentration prevents weak compactness in L1 exampleSuppose a subsequence converged weakly in $L^1$ to $f$. For each $m$, all sufficiently late supports lie inside $B_{1/m}(0)$. Testing against every bounded measurable function supported outside that ball shows that $f=0$ almost everywhere there. Taking the countable union over $m$ proves $f=0$ almost everywhere on ${\mathbb{R}}^n$. But testing against the bounded constant function one gives $\int f={\lim }_j\int f_j=1$, a contradiction.

The same argument applies to every subnet whose indices tend to infinity, so the sequence has no weak cluster point in the ball and the ball is not weakly compact. It converges against smooth compactly supported tests to a point mass, which is not an $L^1$ density. Thus boundedness in $L^1$ does not imply the weak compactness proved in part (b); concentration is the obstruction.